The messy term on the middle is measurable because it consists of countable unions and intersections of
measurable sets. It equals f−1

((b,∞ ])

and so this last set is also measurable. By Lemma 5.1.2, f is
measurable. ■

Observation 5.1.4If f : Ω → ℝ then the above definition of measurability holds with no change.In this case, f never achieves the value ∞. This is actually the case of most interest.

The following theorem is of major significance. I will use this whenever it is convenient.

Theorem 5.1.5Suppose f : ℝ → ℝ is measurable and g : ℝ → ℝ is continuous. Then g ∘ f ismeasurable. Also, if f,g are measurable real valued functions, then their sum is also measurable andreal valued as are all linear combinations of these functions.

Proof:

(g ∘f)

−1

((a,∞ ))

= f−1

( −1 )
g ((a,∞ ))

and by continuity of g, it follows that g−1

((a,∞ ))

is
an open set. Thus it is the disjoint union of countably many open intervals by Theorem 2.7.10.
It follows that f−1

( −1 )
g ((a,∞ ))

is the countable union of measurable sets and is therefore
measurable.

Why is f + g measurable when f,g are real valued measurable functions? This is a little trickier. Let the
rational numbers be

{rn}

n=1∞.

(f + g)−1((a,∞ )) = ∪∞n=1g−1(rn,∞ )∩f −1(a− rn)

It is clear that the expression on the right is contained in

(f + g)

−1

(a,∞ )

. Why are they actually equal?
Suppose ω ∈

(f + g)

−1

(a,∞ )

. Then f

(ω)

+ g

(ω )

> a and there exists rn a rational number smaller than
g

(ω )

such that f

(ω)

+ rn> a. Therefore, ω ∈ g−1

(r,∞ )
n

∩ f−1

(a − r )
n

and so the two sets are
actually equal as claimed. Now by the first part, if f is measurable and a is a real number,
then af is measurable also. Thus linear combinations of measurable functions are measurable.
■

Although the above will usually be enough in this book, it is useful to have a theorem about
measurability of a continuous combination of measurable functions. First note the following.

There is a fundamental theorem about the relationship of simple functions to measurable functions
given in the next theorem.

Definition 5.1.8Let E ∈ℱ for ℱ a σ algebra. Then

{
XE (ω ) ≡ 1 if ω ∈ E
0 if ω ∕∈ E

This is called the indicator function of the set E. Let s :

(Ω, ℱ)

→ ℝ. Then s is a simple function if it is ofthe form

∑n
s (ω ) = ciXEi (ω)
i=1

where Ei∈ℱ and ci∈ ℝ, the Eibeing disjoint. Thus simple functions have finitely many values and aremeasurable. In the next theorem, it will also be assumed that each ci≥ 0.

Each simple function is measurable. This is easily seen as follows. First of all, you can assume the ci are
distinct because if not, you could just replace those Ei which correspond to a single value with their union.
Then if you have any open interval

(a,b)

,

s−1((a,b)) = ∪ {Ei : ci ∈ (a,b)}

and this is measurable because it is the finite union of measurable sets.

If f is bounded, the convergence is actually uniform. Conversely, if f is nonnegative and is the pointwiselimit of such simple functions, then f is measurable.

Proof: Letting I ≡

{ω : f (ω ) = ∞ }

, define

2n
t (ω) = ∑ kX kk+1 (ω)+ 2nX (ω ).
n k=0 n f−1([n ,n-)) I

Then tn(ω) ≤ f(ω) for all ω and limn→∞tn(ω) = f(ω) for all ω. This is because tn

(ω)

= 2n for ω ∈ I and
if f

(ω)

∈ [0,

2n+1
n

), then

0 ≤ f (ω)− tn(ω) ≤ 1. (5.3)
n

(5.3)

Thus whenever ω

∕∈

I, the above inequality will hold for all n large enough. Let

s1 = t1,s2 = max (t1,t2),s3 = max (t1,t2,t3),⋅⋅⋅.

Then the sequence {sn} satisfies 5.1-5.2. Also each sn has finitely many values and is measurable. To see
this, note that

s−1((a,∞ ]) = ∪n t−1 ((a,∞ ]) ∈ ℱ
n k=1k

To verify the last claim, note that in this case the term 2nXI(ω) is not present and for n large enough,
2n∕n is larger than all values of f. Therefore, for all n large enough, 5.3 holds for all ω. Thus the
convergence is uniform.

Now consider the converse assertion. Why is f measurable if it is the pointwise limit of an increasing
sequence simple functions?